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### Course: AP®︎/College Calculus BC > Unit 11

Lesson 3: AP Calculus BC 2011- 2011 Calculus BC free response #1a
- 2011 Calculus BC free response #1 (b & c)
- 2011 Calculus BC free response #1d
- 2011 Calculus BC free response #3a
- 2011 Calculus BC free response #3 (b & c)
- 2011 Calculus BC free response #6a
- 2011 Calculus BC free response #6b
- 2011 Calculus BC free response #6c
- 2011 Calculus BC free response #6d

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# 2011 Calculus BC free response #6d

Lagrange error bound for Taylor Polynomial approximation. Created by Sal Khan.

## Want to join the conversation?

- I assume this may be something not required for Mastery of Integral Calculus at Khan Academy as there were no practice problems or Mastery Challenges on this subject, just a proof this subject given in the video series on Taylor Series. Why are there not practice problems on this subject? Why wouldn't this be required for mastery?(4 votes)
- Khan Academy is an on-going project, so there are not yet Mastery Challenges and/or practice problems available for every topic.

But the ability to work with Taylor series is standard calculus curriculum and is needed for applied and advanced level mathematics.(6 votes)

- Is it possible to evaluate M without looking at the graph?(5 votes)
- i still do not understand why the M=40, when M could also equal 80 or 100 or etc. according to this logic. Why can the M be its set to it corresponding value?(2 votes)
- Why do you have to know the n+1th derivative of a function to bound the error on the of the approximation? I don't understand the relationship, not even after watching Sal's proof videos.(1 vote)
- because the n+1th derivative of an n-degree polynomial is 0. If y = x then dy/dx = 1 ... so dy^2/dx = 0 because the derivative of a constant is 0. You can know that's true without knowing anything else... so it gets used to prove the error bound.(1 vote)

- At 4: 40, why does b have to be greater than a?(1 vote)
- Could you also use the alternating series error bound?(1 vote)
- at around8:49, is it also correct to let M=30 to solve the problem? Or is letting M=40 more correct?(1 vote)
- Can you possibly speak up because my volume is all the way up.(0 votes)

## Video transcript

Part D. Let p sub 4 of x
be the fourth degree Taylor polynomial for f
about x equals 0. Using information
from the graph of y is equal to the absolute value
of the fifth derivative of f of x, shown above-- and that's
the graph right over here, I put it on the side
to save space-- show that the absolute
value of the difference between the polynomial--
the fourth degree polynomial evaluated at 1/4 and the
function evaluated at 1/4 is less than one over 3,000. So this is, it seems like quite
an interesting problem here. So first, let's just think
about what they're asking. They're asking
for the difference between the polynomial and
the function evaluated 1/4. In other words,
they're asking us to-- they're asking us to
bound the error of the function approximation at
x is equal to 1/4. So let me draw a
little graph here just so we can visualize
it a little better. So if that's our y-axis and--
let me draw it like this. So this is our y-axis
and this is our x-axis. And our function, let's
say our function looks something like this. Our polynomial is centered at 0. Our fourth degree
polynomial is centered at 0. And so it'll be equal
to the function at 0. And then it'll become
probably a worse approximation as we get further and
further away from 0. And what they're
saying is, let's bound how bad of
an approximation that is when x is equal to 1/4. So if we say that this is x is
equal to 1/4 right over here, they're saying, look, take the
value of the polynomial at 1/4, subtract from that the value
of the function at 1/4, and then take the
absolute value and then that'll essentially give us the
absolute value of the error. Or that will give us the
absolute value of the remainder or give us the distance between
the two functions there. And they want us to
bound that distance. Now, to do this problem
there is something that you just kind
of have to know. And I've proven it
in another series of videos-- two videos
I think it took me-- in the Khan Academy
Calculus playlist. If you do a search for bounded
or error or approximation or Taylor, you should
probably find it. And that there's a general idea
here which we've proved that if you have some function-- so
I'm going to show the general-- so if you have some function
f of x and that you have some polynomial approximation
of it of degree n--- some polynomial
approximation of degree n. And let's say that this
polynomial approximation is centered at a. In our particular case, a is 0. So that you're centered at a. So we could write-- we
could say this is about a. It's an approximation
about x equals a. We can bound the error. So let me first define
this error function here. Sometimes it's
called a remainder. So we could say, the error
function for the nth degree polynomial about a-- the
n-th degree Taylor polynomial approximation about a-- it
will be a function of x. So for any other x that you
pick, what will this error be? Is equal to the difference
between these two things. You could say it's f of x minus
that or that minus f of x. But let me just
write it this way. This is equal to f of x minus
the polynomial approximation of x. And if you want to take
the absolute value of this, that's equal to
absolute value of that. And if you're taking
absolute values, then you can switch the order. So this is the same thing
as the absolute value of the polynomial approximation
at x minus f of x. So the thing that you just need
to know is, we can bound this. We can bound this if
we know some properties about the n plus 1. So this is the nth
degree approximation. If we know some properties about
the n plus 1th derivative of f. So if we know that
the n plus 1th is the derivative of f is less
than or equal to some maximum value-- and in particular,
the absolute value of the n plus 1th
derivative of f-- is less than or equal to
some-- we could call this some maximum value--
over the interval, over x is in the interval
between a and some b, where b is greater than a. Then we can make the statement--
and this is, once again, something that I
prove in that video. I think I call it the
boundedness of the error remainder for Taylor
approximation. Something like that. Then we know that the
error function at any given x where x is greater than or--
let me say this in particular. Is any given x that's part
of this interval-- where x is part of this
interval between a and b-- is going to be less
than or equal to M times x minus a to the n plus
1th power over n plus 1. So this is the thing
that you really have to know ahead of time to
be able to do this problem. It would be nearly impossible in
the time constraints of the AP exam to prove this
from first principles. I encourage you to
watch that video so you can understand
how this proof happens from first principles. But going into the AP
exam, based on the facts that this did show
up in the 2011 exam, it's probably good to
know this property, to know how to bound it. And this is clearly what they're
asking for because they're giving us-- so we know
what the fourth-- we know what p sub four is. p sub 4 of x-- it's up
to the fourth degree term of the polynomial
approximation of f about 0. In part B, we got all the
way the sixth degree term. So if you want to go just
to the fourth degree term, this is what p sub 4 is. So this is 1 plus x squared over
2 plus x to the fourth over 4. So we know that. And we want to bound this
thing right over here. So if we can find an M-- so
this is the fourth degree polynomial. This is the fourth
degree error that we're talking about in this case. Let me write it this way. This is the nth error
in the general case. But we want to bound
the fourth degree error. So if we want bound the
fourth degree error-- so we can say the fourth
degree error-- at b. In this case, we're
going to say-- or at 1/4 I should say, not b. I never mentioned a b before. I mentioned the b
when I prove it. The fourth degree
error at 1/4 is going to be less than or equal
to-- I did mention a b here so if you take this
to be 1/4, and 1/4 is part of that interval
so you can apply it here. So the fourth degree error,
I guess you could say, at 1/4 is going to be less
than or equal to some M times b-- I should say, 1/4 minus a. a in our particular case
for this problem is a 0. So 1/4 minus is just 0
to the n plus 1th power. So this is 4-- this is the
fourth degree polynomial. n is 4 so the n plus 1 power
is the fifth power. All of that over--
and sorry, this was supposed to be n
plus 1-- all of that over n plus 1 factorial. So this is going to
be over 5 factorial. Don't want to confuse you. This was-- all of that is
over n plus 1 factorial. So all of this is
over 5 factorial. So we can make this statement
if we know that the fifth derivative-- the n
plus 1th derivative-- if we know that the fifth
derivative of f is less than or equal to some value M-- the
absolute value of the fifth derivative of f, I should say---
is less than or equal to some value at M over the interval
x is between 0 and 1/4. Those are our a's and b's
in this particular case. So when you look at
the graph-- and that's exactly what they gave us. They gave us the graph
of the absolute value of the fifth derivative of x. And 1/4 is sitting
right over here. So this is 1/4. We don't know exactly
what we hit over here. This looks like it's roughly,
I don't know, 31 or 32. But we know it's
definitely less than 40 over the interval
between 0 and 1/4. So we can pick M to be 40. We know that this is
less than or equal to 40 over that interval. So we can make this statement. And so we know that this
thing right over here is going to be less
because this over here is the same as this over here. We know this is less than--
let me write it all out. This is-- and I know of people
who got fives on this example without being able to do this. So don't stress out if this
seems really bizarre too. This is really so
they can separate out who happened to
know this, really. So this we can say is
based on this property, we can say is less than
or equal to 40 times 1/4 to the fifth power. So this is 40 times
1/4 to the fifth power. So I could just say divided
by 4 to the fifth power times-- and then the
denominator over here-- we have 5 factorial. I'll just write it out. 5 times 4 times 3 times 2. And we don't have
to write the 1. That doesn't change the value. And now we can simplify this. We can divide the numerator
and the denominator by 4. So this becomes 10,
that just becomes a 1. We can divide it by 5. This becomes a 2,
that becomes a 1. We can divide by 1. That becomes a-- oh,
sorry, we can divide by 2. That becomes a 1,
that becomes a 1. And you say that this
quantity over here is going to be equal to 1. So let me write it. So all of this stuff is less
than or equal to 1 over 4 to the fifth times 3. That's all we have left
in the denominator. And so what is 4 to the fifth? You may or may not know. You can work it out by hand if
you like, but 4 to the fifth is the same thing as
2 to the 10th power because 4 is 2 squared. 2 to the 10th power, if you're
familiar with computer science, is 1,024. But you can work
it out yourself. So this is going to
be-- this is less than or equal to 1,024 times 3. So what is that? That is less than or
equal to 1 over 3. Oh, what is it? 72? 3,000. 1 over 3,072. So this quantity right over
here is less than 1 over 3,072. We know. We know that for a fact. Well, if something is
less than 1 over 3,072, it's definitely going to
be less than 1 over 3,000. 1 over 3,000 is a larger number. It has a smaller denominator. That's definitely less than
or equal to 1 over 3,000. And then we are done. We've proven that this is
less than 1 over 3,000. We've proven that the error at
1/4 is less than 1 over 3,000.